Hey, all. Here is the initial problem:

What it the volume of the solid inside both the regions:

r=a\cos\theta

z^2=a^2-r^2

I set the integral up as follows:

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{a\cos\thet a}2\int_{0}^{\sqrt{a^2-r^2}}r,dz,dr,d\theta

But that doesn't work for some reason, because taking θ from -π/2 to π/2 cancels out, and instead, we have to take θ from π to 2π, like so:

\int_{\pi}^{2\pi}\int_{0}^{a\cos\theta}2\int_{0}^{ \sqrt{a^2-r^2}}r,dz,dr,d\theta

I don't understand why that is.

EDIT: I also don't understand why my LaTeX notation isn't working. Not even simple **** like $\displaystyle r=a$.

EDIT #2: Is it perhaps because both r and θ need to be positive when finding a volume? If so, why would that be true for volume but not area?